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Sagot :
Answer:
28.57
Explanation:
To find the distance the car travels before it stops, we can use the equations of motion for uniformly decelerated motion.
Given:
- Initial velocity, \( u = 20 \) m/s (since the car is slowing down, we take it as positive)
- Acceleration (deceleration), \( a = -7 \) m/s\(^2\) (negative because it's deceleration)
- Final velocity, \( v = 0 \) m/s (since the car stops)
We need to find the distance \( s \).
The equation relating initial velocity, final velocity, acceleration, and distance is:
\[ v^2 = u^2 + 2as \]
Substitute the given values:
\[ 0 = (20)^2 + 2 \cdot (-7) \cdot s \]
Simplify and solve for \( s \):
\[ 0 = 400 - 14s \]
\[ 14s = 400 \]
\[ s = \frac{400}{14} \]
\[ s = 28.57 \text{ meters} \]
So, the car travels approximately \( 28.57 \) meters before it stops.
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